Question

When a reagent $\left(X\right)$ reacts with $F{e}^{3+}$, the solution turns red due to the formation of a compound $\left(Y\right)$. This reagent causes no change in color with $F{e}^{2+}$ in the pure state. Here $\left(X\right)$ and $\left(Y\right)$ are, respectively:

• A. $N{H}_{4}SCN$ and $\left[Fe{\left(SCN\right)}_3\right]$
• B. $K_4\left[Fe{\left(CN\right)}_6\right]$ and $F{e}_4{\left[Fe{\left(CN\right)}_6\right]}_3$
• C. $N{a}_{2}HP{O}_4$ and $FeS{O}_4$
• D. $K_3\left[Fe{\left(CN\right)}_6\right]$ and $K_{2}Fe{\left[Fe{\left(CN\right)}_6\right]}_3$

Answer 1

The correct option is A $N{H}_{4}SCN$ and $\left[Fe{\left(SCN\right)}_3\right]$

The reaction taking place here, is as follows: $FeC{l}_3+3N{H}_{4}SCN\to Fe(SCN{)}_3+3N{H}_{4}Cl$

Here $\left(X\right)$ and $\left(Y\right)$ are, respectively $N{H}_{4}SCN$ and $\left[Fe{\left(SCN\right)}_3\right]$